1. No category

Additional Notes and Solution Manual For:
Thermodynamics
by Enrico Fermi
John L. Weatherwax∗
August 1, 2014
Introduction
Here you’ll find some notes that I wrote up as I worked through this excellent book. I’ve
worked hard to make these notes as good as I can, but I have no illusions that they are perfect.
If you feel that that there is a better way to accomplish or explain an exercise or derivation
presented in these notes; or that one or more of the explanations is unclear, incomplete,
or misleading, please tell me. If you find an error of any kind – technical, grammatical,
typographical, whatever – please tell me that, too. I’ll gladly add to the acknowledgments
in later printings the name of the first person to bring each problem to my attention.
Acknowledgments
Special thanks to (most recent comments are listed first): Neill Warrington, Eduardo Becerra,
Giacomo Caria, Rosario Oliva, Juanico Loran, Glenn Musano, and Don Rintala for helping
improve these notes and solutions.
All comments (no matter how small) are much appreciated. In fact, if you find these notes
useful I would appreciate a contribution in the form of a solution to a problem that is not yet
worked in these notes. Sort of a “take a penny, leave a penny” type of approach. Remember:
pay it forward.
∗
[email protected]
1
Addendums/Clarifications/Derivations
Constraints on Motion in the p-V plane (Page 6)
Note that in general, there is no conditions or requirements on the motion of a piston in
connection with the gas to which it is attached. What this means is that without any other
information the path in (V, p) can be taken arbitrary and simply represents the projection
of points in (V, p, T ) space onto the (V, p) plane. The only requirement is that at every
(V, p) point the gas is assumed to be in equilibrium and must satisfy its equation of state
f (p, V, T ) = 0. In this way the temperature T maybe computed if desired.
If we are told more information about the specific type of path in the (V, p) plane (like in fact
that it is along an isentrope or an isotherm) then the path in the (V, p) plane is completely
specified by its two endpoints.
Equivalence of the Pressure and Volume Formulation of the Work
done during an Isothermal Expansion (Page 9)
Now p1 and p2 are the initial and final pressures for m grams of an ideal gas under an
isothermal expansion. As such evaluating the ideal gas law at the two end points gives
m
p1 V 1 =
RT
(1)
M
m
RT
(2)
p2 V 2 =
M
where T the common temperature. Dividing these two expressions we have
p1 V 1
=1
p2 V 2
(3)
or separating volume and pressure to alternate sides of the equation give
p2
V1
=
.
V2
p1
(4)
It is because of this relationship that we have the equivalence of the two work expressions
given in Fermi Eq. 10 from the book.
The Derivation of the change in Heat Q in terms of the Variables
T and p (Page 20)
The first law of thermodynamic in terms of its “canonical” variables U and V is given by
dU + pdV = dQ .
To express this in terms of the independent variables T and p we first express the differential
of internal energy U in terms of these variables (using standard calculus) as
∂U
∂U
dU =
dp +
dT .
(5)
∂p T
∂T p
In the same way, we next express the differential of V in terms of the variables T and p as
∂V
∂V
dp +
dT .
(6)
dV =
∂p T
∂T p
These two expressions are then inserted into the left hand side of the first law of thermodynamics (replacing dU and dV ) giving
"
#
∂U
∂V
∂V
∂U
dp +
dT + p
dp +
dT = dQ .
(7)
∂p T
∂T p
∂p T
∂T p
Now grouping dp and dT terms we obtain
∂U
∂p
+p
T
∂V
∂p
dp +
T
"
∂U
∂T
+p
p
∂V
∂T
#
dT = dQ .
(8)
p
Which is Fermi Eq. 23.
Expressions Relating the Change in Q to the Specific Heats Cp and
CV (Page 24)
Now Fermi Eq. 30 is
CV dT + pdV = dQ ,
(9)
and taking the differential of the ideal gas law gives (Fermi Eq. 31) or
pdV + V dp = RdT .
(10)
When Eq. 9 is subtracted from Eq. 10 we obtain
CV dT − V dp = dQ − RdT ,
(11)
or grouping all coefficients of dT we obtain
(CV + R)dT − V dp = dQ .
(12)
This is Fermi Eq. 32. Now a transformation at constant pressure has dp = 0 (by definition)
and remembering the definition of Cp
dQ
,
Cp ≡
dT p
we obtain from Eq. 12
Cp =
dQ
dT
= CV + R .
(13)
p
Which is Fermi Eq. 33. As described in the text this same expression can be obtained for a
ideal gas in the following way. Since for an ideal gas the internal energy U is only a function
of only T the partial derivative holding p constant becomes a total derivative
∂U
dU
=
= CV .
(14)
∂T p dT
Because we know that the general relationship between Cp and CV is given by
dU
dV
dQ
=
+p
.
Cp =
dT p
dT p
dT p
Which for an ideal gas and Eq. 14 becomes
∂
RT
R
Cp = CV + p
= CV + p = CV + R .
∂T
p p
p
(15)
(16)
Which again is Fermi Eq. 33. For a monotonic gas with CV = 23 R the above gives
3
5
Cp = R + R = R
2
2
(17)
For a diatomic gas (with CV = 25 R) the above calculation becomes
7
5
Cp = R + R = R .
2
2
From the definition of K =
Cp
CV
(18)
and the fact that Cp = CV + R we see that
K =1+
R
CV
(19)
For a monotonic gas with CV = 23 R we have that
K =1+
R
3
R
2
=1+
5
2
=
3
3
(20)
=1+
7
2
= ,
5
5
(21)
In the exact same way a diatomic gas has
K =1+
for its ratio of specific heats.
R
5
R
2
Constraints on p, V , and T under an Adiabatic Transformations of
an Ideal Gas (Page 25-26)
For an adiabatic transformation dQ = 0 and from Eq. 9 we have
CV dT + pdV = 0 .
(22)
Inserting the ideal gas law pV = RT in the above for p gives a differential expression in terms
of only T and V . From this, the following sequence of manipulations (assuming constant
CV ) integrates this obtaining the following
RT
dV = 0
V
R dV
dT
+
= 0
T
CV V
R
log(T ) +
log(V ) = constant .
CV
CV dT +
TV
Defining K = 1 +
R
CV
R
CV
= constant .
(23)
(24)
(25)
(26)
the above becomes
T V K−1 = constant .
(27)
For an initial temperature, volume configuration denoted by (T0 , V0 ) the above expression
becomes
T V K−1 = T0 V0K−1 .
(28)
For a diatomic gas we have K = 57 , so K − 1 = 25 = 0.4. For an expansion that has its final
volume V twice its initial volume size 2V0 we have
T =
T0
K−1
2
=
T0
≈ 0.75 T0 .
20.4
(29)
To derive the expression of an adiabatic expansion in terms of the variables p and V or p and
T we substitute the ideal gas law in Eq. 27 as follows. From ideal gas law we have pV = RT
or T = pV
when substitute into Eq. 27 gives
R
pV K−1
V
= constant ,
R
(30)
pV K = constant .
(31)
or
This is Fermi Eq. 39 and is the equation between the variables p and V that must be true for
an adiabatic transformation of an ideal gas. In terms of p and T (replacing V with V = RT
p
in Eq. 27) it looks like
(RT )K−1
= constant .
(32)
T
pK−1
or
TK
= const.
(33)
pK−1
which is equivalent to Fermi Eq. 40. Another way of obtaining the same expression can be
obtained by solving for V in Eq. 27 giving
V =
C
1
T K−1
.
Here I will denote constants that don’t need to be further specified by C. Putting this
expression for V into the ideal gas law (pV = RT ) gives
pC
= RT ,
1
T K−1
(34)
or solving for p we obtain
K
1
p = C T 1+ K−1 = C T K−1 ,
(35)
which we recognize as being equivalent to Fermi Eq. 40 again.
A Comparison of the p-V Representations of Isothermal and Adiabatic Transformations (Page 26-27)
= VC for an isothermal transformation. From the previous
In the (V, p) diagram p = RT
V
section we see that an adiabatic transformation in the (V, p) diagram is governed by p = VCK ,
with K the ratio of specific heats. To explicitly determine the slope for each of these
transformations in the (V, p) plane we must evaluate the following derivatives
−C1
dp = 2
(36)
dV isotherm
V
and
−C2
dp = K+1
dV adiabatic V
(37)
We see that the adiabatic transformation will have a steeper slope when
−C1
−C2
> 2 .
K+1
V
V
(38)
In terms of V (assuming C1 ≈ C2 > 0 for simplicity) the above equation is equivalent to the
following sequence of algebraic transformations
−
1
V
K+1
1
<
V
V K+1 >
K +1 >
K >
K+1
1
V2
1
V2
V2
2
1.
> −
(39)
(40)
(41)
(42)
(43)
Since this last equation is true for a general thermodynamical system, all previous manipulations are valid and we conclude that the slope of isothermal curve is steeper in general.
An Example of Adiabatic Expansion in the Atmosphere (Page 27)
The weight of a constant density object in a gravitational field is given by the product of
the gravitational constant, the objects density, and the objects volume. For a fluid in a an
infinitesimal cylindrical slab of height dh and base area A this weight is given by
dW = gρAdh .
To balance this weight, a pressure differential must exist between the top and bottom of the
fluid (which is maintained by the internal pressure) and is given by
dp =
dW
gρAdh
=−
= −ρgdh .
A
A
(44)
We have explicitly introduced a minus sign in the pressure differential under the expectation
that the pressure should decrease dp < 0 as we move up in the atmosphere dh > 0. From
p
Eq. 44 can be written
the ideal gas law in terms of density ρ = M
RT
dp = −
gMp
dh
RT
(45)
To determine the molecular weight M of air we remember that since air is about 78%
Nitrogen, and 21% Oxygen so we can compute the molecular mass of air from its constitute
parts
MN2 = 2(14) = 28 g/mol
MO2 = 2(16) = 32 g/mol .
(46)
(47)
So the average molecular weight of air is given by an appropriately weighted linear combination of that for Nitrogen and Oxygen
Mair = 0.78(28) + 0.21(32) ≈ 28.56 g/mol .
(48)
A more accurate calculation gives Mair ≈ 28.97 g/mol. Fermi Eq. 40 can be written as
T p−
K−1
K
=C,
(49)
assuming the gas is expanding adiabatically. Taking the logarithm of both sides of this
equation we obtain
K−1
log(T ) −
log(p) = log(C) .
(50)
K
Further taking the differential of both sides of the above equation gives
or solving for
dT
T
K − 1 dp
dT
−
= 0,
T
K p
(51)
K − 1 dp
dT
=
.
T
K p
(52)
gives
Replacing the dp term with that from Eq. 45 we obtain
K − 1 gM
dT
=−
dh .
T
K
RT
(53)
This gives for the change in temperate with respect to height the following expression
dT
K − 1 gM
=−
,
dh
K
R
(54)
which is Fermi Eq. 42 in the book. Since air is mostly a diatomic gas we have that K = 57 .
Taking the remaining constants to be their standard CGS values of
g = 980 cm/s2
dyne cm
,
R = 8.314 107
mol K
we obtain
dT
= −9.8 10−5 deg/cm = −9.8 10−5 deg/kilometer .
dh
(55)
(56)
(57)
Postulates of Reversible Thermodynamic Engines (Page 37-39)
The expression
Ltotal = N ′ L′ − NL ,
is represented in two terms, the first N ′ L′ , is the work done by the first engine and the
second NL is the work done by the heat. The total heat absorbed from T2 is given by
Q2,total = N ′ Q′2 − NQ2 ,
while the total given up to temperature T1 , is
Q1,total = N ′ Q′1 − NQ1 .
With the definitions given by Fermi Eq. 47 and Fermi Eq. 48 for (L′ and L) we have
Ltotal =
=
=
=
N ′ L′ − NL
N ′ (Q′2 − Q′1 ) − N(Q2 − Q1 )
N ′ Q′2 − NQ2 − N ′ Q′1 + NQ1
Q2,total − Q1,total .
(58)
(59)
(60)
(61)
If Q2,total = 0, from the above equation we have that Ltotal = −Q1,total which is Fermi Eq. 51.
Now as explained in the book Ltotal ≤ 0, and because of the equivalence between the magnitude of Ltotal and Q1,total we therefore must have Q1,total ≥ 0. Remembering the definition
of Q1,total and Fermi Eq. 49 the following manipulations derive the fact that the efficiency of
a reversible cyclic engine must be greater than that from a non-reversible cyclic engine.
Q1,total
N ′ Q′1
N′ ′
Q
N 1
Q2 ′
Q
Q′2 1
Q2 Q′1
Q2
Q′1
Q1
Q2
Q1
1−
Q2
ηrev.
≥ 0
≥ NQ1
(62)
(63)
≥ Q1
(64)
≥ Q1
(65)
≥ Q1 Q′2
Q′2
≥
Q′1
Q′1
≤
Q′2
Q′1
≥ 1− ′
Q2
≥ ηnon−rev.
(66)
(67)
(68)
(69)
(70)
Carnot Cycles with an Ideal Gas (Pages 42-43)
We will step along each segment of the Carnot cycle, deriving expression that must be
satisfied due to the known nature of the transformation that occurs during the considered
segment. For instance, from the first law of thermodynamics we obtain that the transformation from A to B along an isothermal expansion must satisfy
UB − UA + LAB = QAB ≡ Q2 .
(71)
For an isothermal expansion the work can be explicitly calculated and is found to be
LAB = RT2 log(
VB
)
VA
(72)
Since the assumed substance is an ideal gas where the internal energy is a function of only
temperature we have that UB = UA and thus
Q2 = LAB = RT2 log(
VB
)
VA
(73)
Along the symmetric isothermal contraction we have
UC − UD + LDC = QDC ≡ Q1
(74)
with (by the same reasoning UC = UD ) and the above simplifies as before
Q1 = LDC = RT1 log(
VD
)
VC
(75)
The two paths we have not considered are the two adiabatic ones CA and BD. On the CA
path the adiabatic constraint requires that
T1 VCK−1 = T2 VAK−1 ,
(76)
T2 VBK−1 = T1 VDK−1 .
(77)
and the BD path requires
Dividing these two expressions gives
VC
VD
K−1
=
VA
VB
K−1
(78)
or taking the 1/(K − 1)th root of both sides we obtain
VC
VA
=
.
VD
VB
(79)
RT2 log( VVAB )
Q2
T2
=
,
=
VD
Q1
T1
RT1 log( VC )
(80)
Thus in terms of Q2 and Q1 we have
which expresses the ratio of heat extracted and emitted in terms of the two operating reservoir
temperatures.
The Efficiency of the Carnot Cycle (Pages 43-44)
From Fermi Eq. 43 we have L = Q2 − Q1 which is equivalent to
1=
Q2
L
−
Q1 Q1
(81)
Using the relationship above to express the heat ratio in terms of the reservoir temperatures
we obtain
L
T2
−
(82)
1=
T1 Q1
Solving for Q1 gives
T1
,
T2 − T1
which is Fermi Eq. 60. Now solving the above for L gives
T1 − T2
T2
L=
Q1 = Q1
−1 .
T1
T1
Q1 = L
(83)
(84)
Proof that the Total Entropy Must Increase under Heat Flow (Page
56)
The transfer of an amount Q between bodies A1 and A2 results in a decrease in entropy ∆S1
for body A1 and an increase in entropy of ∆S2 for body A2 . In terms of the heat transferred
at the lower temperature T1 and the higher temperature T2 we have
Q
T1
Q
= − .
T2
∆S1 =
(85)
∆S2
(86)
The total entropy change in the entire system is given by
∆S1 + ∆S2 =
Q
Q(T2 − T1 )
Q
−
=
> 0,
T1 T2
T1 T2
(87)
which can be seen to be positive since T2 > T1 .
A Derivation of the Functional form of the Entropy (Page 58)
Since entropy is additive while probabilities are multiplicative our entropy function f must
satisfy the relation f (xy) = f (x) + f (y). Replaying y with 1 + ǫ this becomes
f (x(1 + ǫ)) = f (x) + f (1 + ǫ) .
(88)
Expanding side of the above in a Taylor series for small ǫ we obtain
f (x) + xǫf ′ (x) +
x2 ǫ2 ′′
f ′′ (1) 2
f (x) + O(ǫ3 ) = f (x) + f (1) + f ′ (1)ǫ +
ǫ + O(ǫ3 )
2
2
(89)
or canceling the common f (x) we obtain
xǫf ′ (x) + O(ǫ2 ) = f (1) + f ′ (1)ǫ + O(ǫ2 ) .
(90)
Matching powers of epsilon on both sides we have f (1) = 0 and xf ′ (x) = f ′ (1) ≡ k or
f ′ (x) =
k
x
(91)
Integrating, we obtain a functional form for the entropy of f (x) = k log(x) + constant.
Explicit expressions for Entropy (Pages 59-61)
The heat dQ received during an infinitesimal transformation is given by
∂U
∂U
dT +
+ p dV ,
dQ =
∂T V
∂V T
(92)
the above becomes
which is Fermi Eq. 79. Since for a reversible transformation dS = dQ
T
dQ
1
∂U
1 ∂U
dS =
dT +
+ p dV
(93)
=
T
T ∂T V
T
∂V T
For an ideal gas the heat received during an infinitesimal transformation of an ideal gas
comes in two parts as dQ = CV dT + pdV where p = RT
, giving
V
dQ = CV dT +
RT
dV
V
which is Fermi Eq. 84. A reversible transformation has dS =
dS =
(94)
dQ
T
giving
CV
R
dT + dV .
T
V
(95)
Assuming that CV a constant (which is true for an ideal gas) we can integrate the above to
obtain
S = CV ln(T ) + R ln(V ) + a .
(96)
Which is Fermi Eq. 86. Since an idea gas has V =
of p and T obtaining
RT
p
we can determine the entropy in terms
S(p, T ) = CV ln(T ) + R ln(
RT
) + a.
p
(97)
When we expanding out the second ln term we get
S(p, T ) = CV ln(T ) + R ln(R) + R ln(T ) − R ln(p) + a
(98)
S(p, T ) = (CV + R) ln(T ) − R ln(p) + a + R ln(R)
= Cp ln(T ) − R ln(p) + a + R ln(R)
(99)
or
which is Fermi Eq. 87. In the general case where the internal energy U is a function of both T
and V (and Cv is not constant), then Fermi Eq. 80 requires (since S is an exact differential)
and we can equate mixed partials
∂
∂
1 ∂U
1
∂U
=
+p
.
(100)
∂V T ∂T V
∂T T
∂V T
evaluating the derivatives of both sides we obtain
2
1 ∂2U
1
∂ U
1
∂p
∂U
.
=
+p
− 2 +
+
T ∂V ∂T
∂V
T
T ∂T ∂V
∂T
Which upon canceling the second derivative term from both sides gives
∂U
∂p
+p=T
∂V
∂T V
or
∂U
∂V
T
=T
∂p
∂T
T
−p
(101)
(102)
(103)
which is Fermi Eq. 88 in the book. Note that all the derivations up to this point have been
independent of the equation of state. If in fact we further specify that our operating medium
is an ideal gas with an equation of state given by the ideal gas law
p=
RT
V
we can evaluate our general expression equation 103. For the ideal gas law this gives
R RT
∂U
=T −
= 0.
(104)
∂V T
V
V
From which we can conclude that U does not depend on T .
Using T and p as independent variables:
Choosing (T, p) as the independent state variables Fermi Eq. 23 gives for dQ
!
∂U
∂V
∂V
∂U
+p
dT +
+p
dp
dQ =
∂T p
∂T p
∂p T
∂p T
(105)
For a reversible path we again have dS = dQ
(for a general path dS ≥ dQ
) and the above
T
T
becomes
!
∂U
∂V
1
∂U
∂V
1
+p
dT +
+p
dp
(106)
dS =
T
∂T p
∂T p
T
∂p T
∂p T
Again since dS must be a perfect differential equating the cross derivatives we have
!
∂
∂
∂U
∂V
∂U
∂V
1
1
+p
=
+p
.
∂p T
∂T p
∂T p
∂T T
∂p T
∂p T
(107)
when we expanding the derivatives above (remembering that in this formulation p is a
constant to T and vice versa) gives
2
∂V
∂ U
∂U
1
∂V
∂2V
1
=
+p
− 2
+
+p
(108)
T ∂p∂T
∂T
∂T ∂p
∂p T
∂p T
T
2
1
∂ U
∂2V
+
(109)
+p
T ∂p∂T
∂T ∂p
or when one cancels common terms, one obtains
∂V
∂V
∂U
= −p
−T
∂p T
∂p T
∂T p
which is Fermi Eq. 89 in the book.
(110)
Using p and V as independent variables:
With p and V as independent variables with Fermi Eq. 24 dQ can be expressed as
"
#
∂U
∂U
dp +
+ p dV = dQ .
∂p V
∂V p
dQ
T
and the above becomes
"
#
1
∂U
1 ∂U
dQ
dp +
+ p dV .
=
dS =
T
T ∂p V
T
∂V p
On a reversible path dS =
(111)
(112)
As in all previous derivations, using the knowledge that dS is a perfect differential we can
equate mixed partials of S obtaining
∂
∂
1 ∂U
1 ∂U
=
+p
(113)
∂V T ∂p
∂p T ∂V
which upon expanding the above gives us
2
∂
1 ∂U
1
∂U
∂ U
1 ∂2U
∂
1
+
=
+p +
+1
∂V T ∂p
T ∂p∂V
∂p T
∂V
T ∂p∂V
or
1
− 2
T
∂T
∂V
p
∂U
∂p
V
1
=− 2
T
∂T
∂V
p
∂U
∂V
+p
p
!
+
1
.
T
(114)
(115)
Solving for T (by multiplying by T 2 ) we have
∂T
T =−
∂V
p
∂U
∂p
+
V
∂T
∂p
V
∂U
∂V
p
+p
!
(116)
Which is Fermi Eq. 90. in the book.
The Derivation of the Clapeyron Equation (Page 65)
With the change in volume dV and dU given by
dV = (v2 (T ) − v1 (T ))dm
dU = (u2 (T ) − u1 (T ))dm
(117)
(118)
where vi and ui are the specific volume and specific internal energy of the substance. The
direct ratio with
u2 − u1 + p(v2 − v1 ) = λ
gives
dU
dV
=
T
u2 (T ) − u1 (T )
λ − p(v2 (T ) − v1 (T ))
λ
=
=
−p
v2 (T ) − v1 (T )
v2 (T ) − v1 (T )
v2 − v1
(119)
Comparing this with equation 104 or
∂U
∂p
=T
−p
∂V T
∂T T
we have by inspection that
T
dp
λ
=
.
dT
v2 − v1
(120)
(121)
Which is Fermi Eq. 94.
An Example using the Clapeyron Equation (Page 66)
Now with the given value of λ
λ = 540cal/gm = 540(4.185 107erg)/gm = 2259 107erg/gm
and since water boils at 100 Centigrade, T = 100 + 273.15 = 373.15K, so that the Clapeyron
equation gives
(2259 107erg/gm)
dp
=
= 3.61 104erg/cm3 K
3
dT
(373.15K)(1677 − 1.043)cm /gm
Remembering the conversion that
1
dyne
= 7.5 10−5cm Hg
2
cm
we have then
dp
cm Hg
= 2.7
.
dT
K
Normally for gas-liquid or a gas-solid interface
v1 ≪ v2
(122)
(123)
Here v1 is the specific volume of the liquid/solid and v2 is the specific volume of gas. Fermi
Eq. 6 (the equation of state for an ideal gas) gives
p
V
1
=
RT .
m
M
(124)
V
is the specific volume or the volume per gram so, since index 2 corresponds
The expression m
RT
to the gas we have v2 = M
, so that the Clapeyron equation becomes
p
λMp
dp
λ
λ
= =
=
RT
dT
T v2
RT 2
T M
p
Which is Fermi Eq. 96 in the book.
(125)
For water vapor over its liquid at the boiling temperature we get (remembering that M =
18gm/mol)
(2260 107erg/gm)(18gm/mol)(1atm)
dp
=
.
(126)
dT
(8.314 J/molK)(373.1K)2
Remembering the unit conversion that 1erg = 10−7 J and 1atm = 1.01 105 Pa, the above
becomes
dp
(2260J/gm)(18gm/mol)(1.01 105 Pa)
3
4 dynes
=
=
3.55
10
Pa/K
=
3.55
10
.
dT
(8.314 J/molK)(373.1K)2
cm2 K
(127)
Which agrees with the number given in the book.
In equation 125, if the heat of vaporization λ can be assumed independent of temperature
then this expression can be integrated giving
λM
p = Ce RT
(128)
As an example of a solid-liquid system consider melting of ice then we are told that
λ = 80cal/cm = 80 × 4.185 107erg/gm = 334.8 107erg/gm .
(129)
so that the Clapeyron equation becomes
λ
335 107erg/gm
dp
=
=
= −1.35 108 erg/cm3 K
3
dT
T (v2 − v1 )
(273.1K)(1.00013 − 1.0907)cm /gm
Remembering that
1
(130)
dyne
= 9.86 10−7atm
cm2
the above becomes
dp
= −1.33 102 atm/K = −133 atm/K
(131)
dT
This increases in pressure by 134 atmospheres lowers the melting point by 1 Kelvin. The
more pressure ice is under the easier it is to melt it.
The Van der Waals Critical State in terms of the constants a and b
(Page 72-73)
A gas that satisfies the Van der Waal equation of state must satisfy
a p + 2 (V − b) = RT .
V
(132)
(V 2 p + a)(V − b) = V 2 RT
V 3 p − bV 2 p + aV − ab = V 2 RT
pV 3 + (−bp − RT )V 2 + aV − ab = 0 .
(133)
(134)
(135)
Given T and p, to obtain a cubic equation in V we multiplying both sides of the above by
V 2 and expanding the product on the left hand side producing
Evaluating the above at the critical pressure p = pc and temperature T = Tc we have
pc V 3 − (bpc + RTc )V 2 + aV − ab = 0 .
(136)
Because the pressure and temperature are evaluated at the critical isotherm, this expression
must have Vc as a root of third order. Mathematically this means that the above should be
represented as pc (V − Vc )3 = 0 for some choice of Vc . Expanding this cube we have
pc V 3 − 3pc V 2 Vc + 3pc V Vc2 − pc Vc3 = 0 .
(137)
Comparing these coefficients to those in Eq. 136 we have
−3pc Vc = −(bpc + RTc )
3Vc2 pc = a
−pc Vc3 = −ab
(138)
(139)
(140)
This is a set of three equations involving three unknowns of (Vc , pc , Tc ). Solving for Vc using
the second and third equation above by dividing the third equation by the second gives
Vc = 3b .
(141)
Inserting this equation into the second equation from 138 we see that pc is given by
pc =
a
a
.
=
2
3Vc
27b2
(142)
while the first equation from 138 then gives for Tc
a
a
3(3b)( 27b
3Vc pc − bpc
2 ) − b( 27b2 )
Tc =
=
=
R
R
a
b
1
3
−
R
1
27
=
8 a
27 Rb
(143)
which is Fermi Eq. 100. As suggested in the text introducing the non-dimensional variables
p
pc
V
=
Vc
T
=
Tc
P =
(144)
V
(145)
T
(146)
we get by substituting the following
p = Ppc
V = V Vc
T = T Tc ,
into the Van der Waals’ equation of state (Fermi Eq. 99)
a p + 2 (V − b) = RT ,
V
(147)
(148)
(149)
(150)
the following
Ppc +
a
2
Vc V 2
(Vc V − b) = RT Tc ,
(151)
or
a
P+
pc Vc2 V 2
RTc
b
=
T .
V −
Vc
V c pc
(152)
Now the factors involving our recently determined critical constants simplify as follows
a
1
a
= 1 =3
=
a
2
9b
pc V c
27b2
3
b
b
1
=
=
Vc
3b
3
8 a
R 27 Rb
8
RTc
=
=
a
V c pc
3b 27b2
3
and we get for equation 152
3
8
1
P+ 2
= T
V −
V
3
3
(153)
which is Fermi Eq. 101.
Expressions for internal energy and entropy in a Van der Waals gas
(Page 73-75)
Using the Van der Waals equation of state expression (Fermi Eq. 99) by solving for the
pressure we obtain
RT
a
.
(154)
p=− 2+
V
V −b
Now for any thermodynamic system the internal energy change with respect to volume is
given by (holding temperature constant)
∂U
∂p
=T
− p.
(155)
∂V T
∂T V
Evaluating this expression for the p = p(V ) relation given by the Van der Waals equation of
state we obtain
∂U
R
a
RT
a
=T
+ 2−
= 2.
(156)
∂V T
V −b
V
V −b
V
Integrating this expression (with respect to V ) gives
U =−
a
+ f (T ) ,
V
(157)
where f (T ) is an arbitrary function of temperature. This is Fermi Eq. 103. To derive the
entropy of a Van der Waals gas we first evaluate the specific heat at constant volume given
by
∂U
∂Q
=
= f ′ (T ) ,
(158)
CV =
∂T V
∂T V
which if we assume that CV is constant, we can integrate with respect to T to obtain
f (T ) = CV T + w .
This expression when put back into the expression for the internal energy U results in
a
U = − + CV T + w .
(159)
V
Now to evaluate the entropy we can perform the following manipulations exactly as in the
book
dQ
dS =
T
1
(dU + pdV )
=
T
1
a
a
1
RT
=
− 2+
dV
(CV dT + 2 dV ) +
T
V
T
V
V −b
a
a
R
dT
+
dV −
dV +
dV
= CV
2
2
T
TV
TV
V −b
dT
R
= CV
+
dT ,
(160)
T
V −b
which when integrated with respect to T gives
S = CV ln(T ) + R ln(V − b) + constant .
(161)
Which is Fermi Eq. 105. Compare this expression with Fermi Eq. 86 the similar expression
for an ideal gas which is
S = CV ln(T ) + R ln(V ) + a
Now along an adiabatic transformation by definition dQ = 0, equivalently dS = 0, or
S = constant and the above expression can be manipulated as follows
CV ln(T ) + R ln(V − b) = C1
R
ln(V − b) = C2
ln(T ) +
CV
T (V − b)R/CV = C3
(162)
where C1 , C2 , and C3 are all constants. This expression can be recognized as Fermi Eq. 106.
The derivation of the isochore of Van’t Hoff (Page 81)
The derivation given in the book proceeds smoothly until about the middle of the page
where the statement dF (A)/dT = −S(A) is made. This expression can be derived as
follows. Considering the definition of the free energy F (A) as F = U − T S we have that the
temperature derivative of F given by
dF (A)
dU(A)
dS(A)
=
−T
− S(A)
dT
dT
dT
1
(dU(A) − T dS(A)) − S(A)
=
dT
= −S(A)
where we have used the fact that from the first and second law of thermodynamics dU =
dQ − dW = T dS − dW or dU − T dS = −dW = 0 if no work is done.
Since S(A) = (U − F )/T from the definition of the free energy the temperature derivative
above can be written in terms of the free energy and internal energy as
dF (A)
F (A) U(A)
=
−
.
dT
T
T
To relate this to the derivative of work with respect to temperature and derive the isochore
of Van’t Hoff recall Fermi Eq. 115 or
dL
dF (A) dF (B)
=
−
,
dT
dT
dT
we can use the expression above to replace the derivatives of the free energy with respect to
temperature with expressions involving the free energy itself as
dF (A) dF (B)
F (A) − U(A) F (B) − U(B)
dL
=
−
=
−
.
dT
dT
dT
T
T
multiplying both sides by T we obtain
T
dL
= F (A) − U(A) − (F (B) − U(B)) = F (A) − F (B) − (U(A) − U(B)) = L + ∆U
dT
where we have used Fermi Eq. 114 of L = F (A) − F (B). Thus in summary we have derived
−∆U = L − T
dL
dT
(163)
which is Fermi Eq. 117 or the isochore of Van’t Hoff.
WWX: I have not finished this section ... start
WWX: page 2 of the second set of scanned notes
From Fermi Eq. 112 we have L ≤ −∆F
pdV =
so
−p =
∂F
−
∂V
∂F
∂V
(164)
T
(165)
T
Since F = U − ST , for an ideal gas we get that
F = Cv T + W − (Cv T + R log(V ) + a)T
= Cv T + W − T (Cp log(T ) − R log(p) + a + R log(R))
(166)
(167)
Derivation of the thermodynamic potential at constant pressure
(Page 82-83)
L ≤ −∆F gives pV (B) − pV (A) ≤ F (A) − F (B). Now defining
Φ = F + pV = U − T S + pV
(168)
pV (B) + F (B) ≤ pV (A) + F (A)
(169)
so that the above becomes
giving Φ(B) ≤ Φ(A). From the book: To find an equilibrium state for a system that has
pressure and temperature we look for the minimum of the Gibbs Free energy Φ. For a general
system, find what ... thermodynamic potential be it F − U − T S, Φ = U − T S + pV now
since Φ = U − T S + pV we have that
∂Φ
∂U
∂S
∂V
=
−T
+V +p
(170)
∂p T
∂p T
∂p T
∂p T
since dQ = T dS = dU + pdV ,
T
∂S
∂p
=
T
∂U
∂p
(171)
T
putting this into the above we have
∂Φ
∂U
∂U
∂V
∂V
=
−
−p
+V +p
=V
∂p T
∂p T
∂p T
∂p T
∂p T
which is Fermi Eq. 123. In the same way
∂U
∂S
∂V
∂Φ
=
−S−T
+p
∂T p
∂T p
∂T p
∂T p
T dS = dU + pdV
so that
T
∂S
∂T
p
=
∂U
∂T
T
+p
(172)
(173)
(174)
∂V
∂T
(175)
p
when put in the above becomes
∂Φ
∂U
∂U
∂V
∂V
=
−S−
−p
+ p
∂T p
∂T p
∂T p
∂T p
∂T p
(176)
Φ = U − T S + pV . Now T is the same for both vapor and liquid phases and P is the same
for both vapor and the liquid at least at the surface or in zero gravity? Now Φ = Φ1 + Φ2 ,
i.e. on the space shuttle or with very little liquid. pV = RT and
Φ = m1 φ1 (T ) + m2 φ2 (T )
(177)
Let m1 → m1 + dm, then Φ becomes
Φ′ = (m1 + dm1 )φ1 (T ) + (m2 − dm1 )φ2 (T ) = Φ + dm1 (φ1 − φ2 )
(178)
As Φ was at a minimum. In order that the value of Φ′ not change from that of Φ, we must
have φ1 = φ2 , which is
U1 − T S1 + pV1 = U2 − T S2 + pV2
(179)
or
U2 − U1 − T (S2 − S1 ) + p(V2 − V1 ) = 0
(180)
taking the partial with respect to T we have
d
d
dp
d
(U2 − U1 ) − (S2 − S1 ) − T
(S2 − S1 ) +
(V2 − V1 ) + p (V2 − V1 ) = 0
dT
dT
dT
dT
dU
dv
ds
=
+p
dT
dT
dT
dU
dS
dV
−T
+p
=0
dT
dT
dT
dp
(V2 − V1 ) = 0
−(S2 − S1 ) +
dT
(183)
λ
dp
=
dT
T (V2 − V1 )
(185)
T
with S2 − S1 = Tλ , we obtain
Φ =
=
=
=
(181)
U − T S + pV
cV T + W − T (Cp log(T ) − R log(p) + a + R log(R)) + pV
cV T + W + pV − T (Cp log(T ) − R log(p) + a + R log(R))
cV T + W + RT − T (Cp log(T ) − R log(p) + a + R log(R))
(182)
(184)
(186)
(187)
(188)
(189)
since CV + R = Cp the above becomes
Φ = Cp T + W − T (Cp log(T ) − R log(p) + a + R log(R))
dF (A)
dU
dS
=
−T
− S(A)
dT
dT
dT
1
(dU − T dS) − S(A)
=
dT
= −S(A)
(190)
(191)
(192)
(193)
if no work is done? Since
dU = dQ − dW = T dS − dW
(194)
so we have that dU − T dS = −dW = 0 if there is no work done.
F (A) U(A)
dF (A)
=
−
dT
T
T
(195)
, now Fermi Eq. 114 is that L = F (A) − F (B)
since F = U − ST , we have that −S = F −U
T
and
dF (A) dF (B)
dL
=
−
(196)
dT
dT
dT
then
dL
dF (A)
dF (B)
T
=T
−T
(197)
dT
dT
dT
each term in the above can be replaced as
T
dL
= F (A) − U(A) − (F (B) − U(B)) = F (A) − F (B) − (U(A) − U(B)) = L + ∆U (198)
dT
so we have
T
dL
= L + ∆U
dT
(199)
giving
dL
dT
which is Fermi Eq. 117. From Fermi Eq. 112 we have L ≤ −∆F
∂F pdV = −
∂V T
−∆U = L − T
so
−p =
∂F
∂V
(200)
(201)
(202)
T
Since F = U − ST , for an ideal gas we get that
F = Cv T + W − (Cv T + R log(V ) + a)T
= Cv T + W − T (Cp log(T ) − R log(p) + a + R log(R))
(203)
(204)
L ≤ −∆F gives pV (B) − pV (A) ≤ F (A) − F (B). Now defining
Φ = F + pV = U − T S + pV
(205)
pV (B) + F (B) ≤ pV (A) + F (A)
(206)
so that the above becomes
giving Φ(B) ≤ Φ(A). From the book: To find an equilibrium state for a system that has
pressure and temperature we look for the minimum of the Gibbs Free energy Φ. For a general
system, find what ... thermodynamic potential be it F − U − T S, Φ = U − T S + pV now
since Φ = U − T S + pV we have that
∂Φ
∂U
∂S
∂V
=
−T
+V +p
(207)
∂p T
∂p T
∂p T
∂p T
since dQ = T dS = dU + pdV ,
T
∂S
∂p
=
T
∂U
∂p
T
(208)
putting this into the above we have
∂U ∂U ∂V ∂V ∂Φ =
−
−p
+V +p
=V
∂p T
∂p T
∂p T
∂p T
∂p T
which is Fermi Eq. 123. In the same way
∂Φ ∂U ∂S ∂V =
−S−T
+p
∂T p
∂T p
∂T p
∂T p
T dS = dU + pdV
so that
∂U ∂V ∂S =
+p
T
∂T p
∂T T
∂T p
when put in the above becomes
∂U ∂U ∂V ∂V ∂Φ =
−S−
−p
+p
∂T p
∂T p
∂T p
∂T p
∂T p
(209)
(210)
(211)
(212)
(213)
Φ = U − T S + pV . Now T is the same for both vapor and liquid phases and P is the same
for both vapor and the liquid at least at the surface or in zero gravity? Now Φ = Φ1 + Φ2 ,
i.e. on the space shuttle or with very little liquid. pV = RT and
Φ = m1 φ1 (T ) + m2 φ2 (T )
(214)
Let m1 → m1 + dm, then Φ becomes
Φ′ = (m1 + dm1 )φ1 (T ) + (m2 − dm1 )φ2 (T ) = Φ + dm1 (φ1 − φ2 )
(215)
As Φ was at a minimum. In order that the value of Φ′ not change from that of Φ, we must
have φ1 = φ2 , which is
U1 − T S1 + pV1 = U2 − T S2 + pV2
(216)
or
U2 − U1 − T (S2 − S1 ) + p(V2 − V1 ) = 0
(217)
taking the partial with respect to T we have
d
d
dp
d
(U2 − U1 ) − (S2 − S1 ) − T
(S2 − S1 ) +
(V2 − V1 ) + p (V2 − V1 ) = 0
dT
dT
dT
dT
ds
dU
dv
=
+p
dT
dT
dT
dS
dV
dU
−T
+p
=0
dT
dT
dT
dp
−(S2 − S1 ) +
(V2 − V1 ) = 0
dT
(220)
dp
λ
=
dT
T (V2 − V1 )
(222)
T
with S2 − S1 = Tλ , we obtain
(218)
(219)
(221)
Φ =
=
=
=
U − T S + pV
cV T + W − T (Cp log(T ) − R log(p) + a + R log(R)) + pV
cV T + W + pV − T (Cp log(T ) − R log(p) + a + R log(R))
cV T + W + RT − T (Cp log(T ) − R log(p) + a + R log(R))
(223)
(224)
(225)
(226)
since CV + R = Cp the above becomes
Φ = Cp T + W − T (Cp log(T ) − R log(p) + a + R log(R))
(227)
Φi = Φi (T, p, mi,1 , mi,2 , mi,3 , . . . , mi,n )
(228)
Φi (T, p, kmi,1 , kmi,2 , kmi,3 , . . . , kmi,n ) = kΦi (T, p, mi1 , mi2 , . . . , min )
(229)
which is homogeneous of degree 1.
k˙ = mi,k − δm
(230)
mj,k′ = mj,k + δm
(231)
δΦ = δΦi + δΦj
∂Φj
∂Φi
(−δm) +
δm
=
∂mi,k
δmj,k
∂Φi
∂Φj
δm −
δm = 0
=
∂mj,k
∂mi,k
(232)
∂Φj
∂Φi
=
∂mj,k
∂mi,k
(233)
(234)
(235)
How many equations like this do we have? f phases and n components, since each component
n of them in a given phase i can go to any of the other f − 1 phase (nothing happens if it
goes to itself. We have n(f − 1) equations for equilibrium. Each ∂Φ
depends only on ratios
∂·
i
of the mk,i . The number of ratios there are like this are n − 1. Then for all ∂Φ
is f (n − 1)
∂·
with T and p we have f (n − 1) + 2 variables. Let v be the number of unknowns minus the
number of equation. Then
v = (n − 1)f + 2 − n(f − 1)
= 2+n−f
(236)
(237)
which is Fermi Eq. 131. Now example #1 we have
v = 2+1−1 =2
(238)
v = 2+2−1 =3
(239)
Now example #2 we have
Now example #3 is 2 phases solid and liquid has f = 2 and one component so n = 1 so we
have
v =2+n−f =2+1−2= 1
(240)
Now in example #4 we have n = 1 and f = 3 so we have
v =2+n−f =2+1−3= 0
(241)
we have L = ev. Power exerted by DC current is given by P = V I with V the voltage and
I the current. Now current is charge per unit time. Thus the work is voltage times charge.
U(T, e) = U(T ) − eU(T )
(242)
∆U = −eU(T )
(243)
isochore of Vanft Hoff
dL
= −∆U
(244)
dT
d(eV )
eV − T
= eU
(245)
dT
dv
=u
(246)
v−T
dT
which is Fermi Eq. 134. e = Cv(T ) and dL = 12 dCV 2 (T ) energy of an isolated capacitor is
given by
1 e2
(247)
2C
1 e2
1 e2
)=
dC
(248)
dL = −d(
2C
2 C2
L−T
But e = Cv and dL =
1 C 2 v2
dC
2 C2
= 21 v 2 dC
U(T, e) = U(T ) − eu(T ) = U(T ) − Cv(T )u(T )
(249)
plus energy now present in condenser, heat released/absorbed when
1
1
dQ = dU + dL = d(U(T ) − CV (T )u(T ) + Cv 2 (T )) + dCv 2
2
2
dU
dv
dU
dv
= dT
− C u − Cv
+ Cv
dT
dT
dT
dT
1
1
+ dC −vu + v 2 + v 2
2
2
dQ
dT dU
du
dv
dv
dS =
=
− Cv
− Cu
+ Cv
T
T
dT
dT
dT
dT
dC 2
+
v − vu
T
∂
∂C
"
dU
dT
− CV
1
T
dU
dT
dv
− Cu dT
+ CV
T
dV
dT
#
∂
=
∂T
dU
dV
dv
−v
−u
+v
dT
dT
dT
v 2 − vu
T
(250)
(251)
(252)
(253)
(254)
(255)
(256)
= −
v 2 − vu 2vv ′ − v ′ u − vu′
+
T2
T
(257)
which gives
−
v du
u dv
v dv
2v dv
u dv
v du
v2
vu
−
+
=
−
−
− 2+ 2
T dT
T dT
T dT
T dT
T dT
T dT
T
T
vu
v dv
v2
− 2 =
2
T
T
T dT
v
u
dv
− =
T
T
dT
which gives
v−T
dv
=u
dT
(258)
(259)
(260)
(261)
which is Fermi Eq. 134.
End page 10 of second set of scanned notes
begin page 13 of second set of scanned notes
In Fermi Eq. 136 if k(T ) ≪ 1 then
[A1 ]n1 [A2 ]n2 · · · [Ar ]nr = k(T )[β1 ]m1 [β2 ]m2 · · · [βs ]ms
(262)
Generally we have that [A1 ]n1 [A2 ]n2 · · · [Ar ]nr ≪ 1, so the reaction is shifted to the right.
Now the definition of [A] is the number of moles of A divided by the volume. So [A]M is
the mass of A divided by the volume.
LI = −RT
n
X
ni
(263)
i=1
p1 V1 = n1 RT
(264)
since p ∝ Vn RT we have that p ∝ [H]RT , so the pressure is proportional to the mole
concentration. To insure that we get m1 moles in the first cylinder which we can extend to
+∞ we must require that
m1
V1 [β1 ] = m1 ⇒ V1 =
(265)
[β1 ]
where [β1 ] concentration of elements β1 .
L = LI + LII
!
n
s
X
X
ni = −∆F
= RT
mj −
j=1
(266)
(267)
i=1
as this process is isothermal and reversible. We have that
1
) + a1 }
[A]
= n1 {Cv1 T + w1 − T (Cv1 log(T ) − R log([A]) + a1 }
F1 = n1 {Cv1 T + w1 − T (Cv1 log(T ) + R log(
(268)
(269)
(270)
The Free energy initially is given by
r
X
FI =
ni {Cvi T + wi − T (Cvi log(T ) − R log([Ai ]) + ai }
(271)
mj {Cv′ j T + wj′ − T (Cv′ j log(T ) − R log([Bj ]) + a′j }
(272)
i=1
FF =
s
X
j=1
FI − FF = L becomes
!
s
n
n
s
X
X
X
X
RT
mj −
ni =
ni {Cvi T +wi −T (Cvi log(T )−R log([Ai ])+ai }−
mj {Cvi T +wi −T (Cvi log(T )
i=1
j=1
i=1
j=1
(273)
so we have then that
RT
s
X
mj −
j=1
r
X
i=1
ni
!
=
r
X
i=1
s
X
mj Cv′ j T +mj wj′ −m
ni Cvi T +ni wi −ni T Cvi log(T )+ni RT log([Ai ])+ni ai T −
j=1
(274)
dividing by RT we have
s
X
mj −
j=1
r
X
ni =
i=1
r
X
ni Cv
i
i=1
+
R
+
s
X
mj Cv′ j
j=1
R
ni wi ni Cvi log(T )
ni ai
−
+ ni log([Ai ]) +
RT
R
R
+
(275)
mj Cv′ j log(T )
mj wj′
mj aj
−
+ mj log([Bj ]) +
(276)
RT
R
R
so we have that
r
X
ni log([Ai ]) −
i=1
s
X
mj log([Bj ]) = −
j=1
r
X
ni Cv
i
i=1
+
RT
s
X
mj Cv′ j
j=1
R
ni wi ni Cvi log(T ) ni ai
−
+
+ n(277)
i
RT
R
R
+
mj Cv′ j log(T ) mj a′j
mj wj′
+
−
+
+(278)
mj
RT
R
R
which gives
[A1 ]n1 [A2 ]n2 · · · [Ar ]nr
log(
[B1 ]m1 [B2 ]m2 · · · [Br ]mr
r
s
′
X
ni Cvi log(T ) X mj Cvj log(T )
=
−
R
R
i=1
j=1
+ −
r
X
ni wi
i=1
+ −
j=1
RT
+
mj aj
+ mj
R
j=1
r
X
ni Cv
i
i=1
s
X
mj Cv′ j
+
RT
RT
s
X
mj wj′
+
RT
ni ai
+ ni
R
(279)
(280)
(281)
(282)
Therefore we have that
T
log(· · · ) = log(
T
−
r
X
i=1
or
log(· · · ) = log(T
n1 Cv1
R
m1 Cv1
R
T
T
n2 Cv2
R
T
m2 Cv2
R
T
n3 Cv3
R
nr Cvr
R
···T
m3 Cv3
R
···T
)−
ms Cvs
R
r
X
i=1
s
X
mj wj′
ni wi
log(exp(
)) +
log(exp(
))
RT
RT
j=1
(283)
s
X
mj Cv′ j
mi a′j
ni Cvi ni ai
log(exp(
+
+ ni )) +
log(exp(
+
+ mj ))
R
R
R
R
j=1
1
R
Pr
1
i=1 ni Cvi − R
Ps
′
j=1 mj Cvj
Taking exponentials we have
1
[A1 ]n1 [A2 ]n2 · · · [Ar ]nr
= exp(
m
m
m
s
1
2
[B1 ] [B2 ] · · · [Bs ]
R
which is times
T
P
P
mj Cv′
mj a′
mj w ′
exp( sj=1 R j + R j + mj )
exp( sj=1 RT j )
P
)+log(
)
)+log(
P ni Cvi ni ai
i wi
)
exp( ri=1 nRT
+ R + ni )
exp( ri=1 RT
(285)
( s
X
′
mj (R +
Cv′ j
r
X
+ aj ) −
j=1
1
R
(284)
P
r
)
ni (R + Cvi + ai ) ) (286)
i=1
i=1
ni Cvi −
Ps
j=1
mj Cv′ j
(287)
which is times the following
1
exp(−
RT
r
X
i=1
ni wi −
s
X
mj wj′
j=1
!
(288)
which gives Fermi Eq. 139. This gives pi V = ni RT , so pi = niVRT with p =
Since F = U − T S and Φ = U − T S + pV the Free energy is given by
P
i
pi =
RT
V
P
i
ni
Cv1 T +W1 −T (Cv1 log(T )−R log([A1 ])+a1 )+V [A1 ](Cv1 T +W1 −T (Cv1 log(T )−R log([A1 ])+a1 ))
(289)
so
F =V
r
X
i=1
[Ai ](Cvi T +Wi −T (Cvi log(T )−R log([Ai ])+ai ))+V
s
X
[Bj ](Cv′ j T +Wi′ −T (Cv′ j log(T )−R log([Bj ])+
j=1
(290)
Which is Fermi Eq. 140. Thus the variation on F is given by
∂F
∂F
∂F
∂F
∂F
∂F
∂F
∂F
ǫn1 −
ǫn2 −
ǫn3 −. . .−
ǫnr ++
ǫm1 +
ǫm2 +
ǫm3 +. . .+
ǫms
∂[A1 ]
∂[A2 ]
∂[A3 ]
∂[Ar ]
∂[B1 ]
∂[B2 ]
∂[B3 ]
∂[Bs ]
(291)
which gives
δF = −
∂F
∂F
∂F
∂F
∂F
∂F
∂F
∂F
[A1 ]−
[A2 ]−
[A3 ]−. . .−
[Ar ]++
[B1 ]+
[B2 ]+
[B3 ]+. . .+
[Bs ] = 0
∂[A1 ]
∂[A2 ]
∂[A3 ]
∂[Ar ]
∂[B1 ]
∂[B2 ]
∂[B3 ]
∂[Bs ]
(292)
From Fermi Eq. 140. we have
−
r
r
X
X
TR
∂F
=V
(Cvi + Wi − T (Cvi log(T ) − R log([Ai ]) + ai ) + V
[Ai ]
= V T R (293)
∂[Ai ]
[Ai ]
i=1
i=1
This is incorrect. You are taking the derivative of the concentration [Ai ] not the summation
variable [Ai ]. The above is taking the derivative of the summation variable index [Ai ] we get
RT
∂F
= V {Cvi T + wi − T (Cvi log(T ) − R log([Ai ]) + ai }+V [Ai ]
= V {Cvi T + wi − T (Cvi log(T ) − R lo
∂[Ai ]
[Ai ]
(294)
In the same way
∂F
(295)
∂[Bi ]
Then
δF = −[A1 ]V (Cv1 T +w1 −T (Cv1 log(T )−R log([Ai ])+ai )+RT )−[A2 ]V (Cv2 T +w2 −T (Cv2 log(T )−R log([Ai ])
(296)
or
−
r
X
ni {Cvi T + wi − T (Cvi log(T ) − R log([Ai ]) + ai ) + RT }+
i=1
s
X
j=1
∆U =
s
X
mj (Cv′ j T + wj′ ) −
j=1
H = −∆U =
r
X
n
mj Cv′ j T + wj′ − T (Cv′ j log(T ) − R log(
(297)
ni (Cvi T + wi )
(298)
i=1
r
X
ni (Cvi T + wi ) −
s
X
wj (Cv′ j T + wj′ )
(299)
j=1
i=1
r
X
d log(k(T ))
1
=
dT
TR
Cvi ni −
i=1
To exclude the logrithm derivative of 141 write
s
X
j=1
Cv′ j mj
!
+
1
RT 2
(300)
C3
k(T ) = C1 T C2 e− RT
(301)
so that
log(k(T )) = log(C1 ) + C2 log(T ) −
C3
C3
log(e) = log(C1 ) + C2 log(T ) −
RT
RT
(302)
so that
1
RT
r
X
i=1
Cvi ni −
s
X
j=1
Cv′ j mj
!
+
1
RT 2
d log(k(T )
C2
C3
=
+
dT !
T
RT 2
r
s
X
X
ni wi −
mj wj′
i=1
j=1
=
=
1
RT 2
H
RT 2
r
X
i=1
T Cvi ni + ni wi −
s
X
j=1
T Cv′ j m
Note that H = H(T ) thus we can see the temperature dependence. If n1 + n2 + . . . + nr <
m1 + m2 + . . . ms , then shifting the equation to the right increases the pressure.
[A1 ]n1 [A2 ]n2 · · · [Ar ]nr
= k(T )
[B1 ]n1 [B2 ]n2 · · · [Bs ]ns
(307)
Compress the system shrinking V causes the concentrations [A] and [B] to increase. Because
of the concentration inequality n1 + n2 + . . . + nr < m1 + m2 + . . . ms the left hand side
decreases to prevent this [A] increases by the [B] stays the same which implies that the
reaction shifts towards the reactants.
Problem 1
2A → A which gives
[A]2
= k(T )
[A2 ]
(308)
is the equation of the law of mass action. Here [A] is in units of number of moles per the
volume and we are told that our reaction constant k is given by k(18C) = 1.7 10−4. Dalton’s
law of partial pressure says that p = 1 atm = pA + pA2 with pA the partial pressure of the A
species and pA2 the partial pressure of the A2 species. Then we have that
pA =
nA RT
= [A]RT
V
(309)
and
pA2 = [A2 ]RT
(310)
p = ([A] + [A2 ])RT
(311)
so
so [A]2 = k(T )[A2 ] which when put in above we have
p = ([A] +
or
[A]2
)RT
k(T )
(312)
p
[A]2
+ [A] −
=0
k(T )
RT
(313)
which is a quadratic equation for [A]. Since
1.01 105 Pa
41.72J//m3
p
=
=
= 41.72mol/m3
RT
(8.314J/molK)(273.15 + 18K)
J/mol
so we have for [A] the following
q
1
−1 ± 1 − 4 k(T
)
[A] =
1
2 k(T
)
−P
RT
−1 ±
=
q
1+
2
k(T )
4p
k(T )RT
= 8.413 10−2mol/V
(314)
(315)
so the concentration of [A2 ]
[A2 ] =
RT
− [A] = 41.72 − 8.413 10−2 = 41.64 mol/m3
p
(316)
we can check this by considering
[A]2
= 1.6997 10−4
[A2 ]
(317)
since we are asked for the percentage of A we remember that nA = [A]V and nA2 = [A2 ]V
[A]
A
= [A]V[A]V
= [A]+[A
= 2.01 10−3 = 0.201
so the percentage of A is given by = nAn+n
+[A2 ]V
2]
A
2
Problem 2
We have H = 50000cal/mol degree of dissociation a H > 0 which means that this is an
exerthermal reaction expect raising the temperature to shift the reaction towards the left
and we should have an increase in the concentration of A. If I write the equation for the
[A]2
chemical reaction in the other order, I would flip the concentration ratio [A
but would not
2]
change the term e−H/RT which would give a different equation. Where is the inconsistency.
Would be then than H should be given as the heat of reaction from the left to right. Which
if we switch the order of the chemical equation we switch the sign of H.
d log(k(T ))
H
=
dT
RT 2
so
log(k(T )) = C1 −
so that
H
RT
H
k(T ) = C2 e− RT
therefore we have
H
[A]2
= C2 e− RT = k(T )
[A2 ]
(318)
(319)
(320)
(321)
What is C2 ? We know that k(T = 18 C = 291.15) = 1.7 10−4 so that
RT = (8.314 J/molK)(291.15K) = 2.42 103J/mol
since we know that 1J = 0.2388cal we have that RT = 5.78 103cal/mol, so that
H
50000cal/mol
=
= 8.64 101
RT
5.78 103cal/mol
(322)
C2 exp(−8.64 101 ) = 1.7 10−4
(323)
which gives C2 = exp(8.64 101) 1.7 10−4 = 5.66 1033, which is huge. Therefore k(T = 18 C) =
H
) = 2.07 10−4. Now to find the percentage of A we remember that p =
5.66 1033 exp(− RT
pA + pA2 = [A]RT + [A2 ]RT
[A] + [A2 ] =
P
RT
and [A]2 = k(T )[A2 ]
(324)
[A] +
p
[A]2
p
[A]2
=
⇒
+ [A] −
=0
k(T )
RT
k(T )
RT
−1 ±
r
(325)
solving for [A] we have
[A] =
1
k(T )
1−4
1
2 k(T
)
−P
RT
= 9.26 10−2 > 8.41 10−2
(326)
so the concentration of species A2 is given by
[A]2
⇒ [A2 ] = 4.14 101 < 4.16 101
[A2 ] =
k(T )
(327)
so the percentage A is given by 2.22 10−3 or 0.22 percent which has increased the concentration of A as expected.
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Problem Solutions
Chapter 1 (Thermodynamic Systems)
Problem 1
Since the gas is expanding against a constant atmosphere (at pressure p) the total work L
done by the gas can be written as
Z V2
pdV = p(V2 − V1 )
(328)
L=
V1
Using the numbers given in the text we have
L = 2.34 atm (4.01 − 3.12) liters = 2.082 atm liters
(329)
To convert this result into CGS units (centimeters, grams, and seconds) we remember that
1 atm = 1.01 105 Pa = 1.01 105
1 liter = 10−3 m3
1 N = 105 dyne
1 m = 100 cm
N
m2
(330)
(331)
(332)
(333)
Thus we have
N
× 10−3 m3
m2
= 1.01 102N m
= 1.01 102 × 105 dyne × 102 cm
= 1.01 109dyne cm .
1 atm liter = 1.01 105
(334)
(335)
(336)
(337)
After unit conversion, our total work L is given by
L = 2.103 102N m = 210.3 N m = 2.103 109dyne cm
(338)
Problem 2
Since we are told that the gas is hydrogen (H2 ), looking to the periodic table gives us that
hydrogen’s mass per unit mole (molecular weight) is given by M = 2g/mole. The rest of the
problem provides the gas’s mass m, volume V , and temperature T as
m = 30 g
T = 18◦ C = 18 + 273.15 = 291.15 K
V = 1 m3
(339)
(340)
(341)
Then one application of the ideal gas law
pV =
m
RT
M
gives
p=
m
1
30g
J
1
J
RT
=
× 8.314
× 291.15 K ×
= 3.63 104 3
3
M
V
2g/mole
mole K
1m
m
(342)
Since 1J = 1N m, the above simplifies to
p = 3.64 104
N
= 3.64 104 Pa .
m2
(343)
To convert to atmospheres, requires the conversion to atmospheres from Pascals given by
1 Pa =
1
atm = 9.9 10−6 atm .
1.01 105
With this, the pressure in atmospheres becomes
p = 3.59 10−1atm .
(344)
Problem 3 (calculate the density and specific volume of nitrogen)
Nitrogen exists as a diatomic molecule (N2 ) under standard conditions and thus has a molecular weight of
M = 2 × 14 grams/mole = 28 grams/mole .
The idea gas law gives for the density ρ
Mp
m
=
.
(345)
V
RT
In the problem, we are assuming that the Nitrogen is at 0 Celsius or 273.15 K. As such, the
density will be completely determined once we have specified a pressure. Since none is given
in this problem we will assume that atmospheric conditions apply and that the pressure is
1 atm = 1.01 105 Pa. With this assumptions the density is given by
ρ≡
ρ =
=
=
Since the specific volume is
that
(28 grams/mole) (1.01 105Pa)
(8.314 J/mole K) (273.15 K)
(10−3 kg)(kg/ms2 )
grams Pa
= 1244
1244
J
kg m2 /s2
kg
g
1.244 3 = 1.244 10−3 3 .
m
cm
defined to be the reciprocal of the density, we have numerically
v=
m3
cm3
1
= 0.8038585
= 803.8585
.
ρ
kg
g
(346)
Problem 4
From Fermi Eq. 9 in the book we have the work for an isothermal expansion is given by
L=
m
V2
m
p1
RT log( ) =
RT log( ) .
M
V1
M
p2
(347)
For this problem the provided inputs are
m
M
R
T
p1
p2
=
=
=
=
=
=
10 g
32.0 g/mol
8.314J/molK
20 C = 293.15 K
1 atm
0.3 atm
with these inputs the expression for work above becomes
10g
J
1
L=
8.314
(293.15 K) log( )
32g/mol
molK
0.3
(348)
which simplify to
L = 916.9J .
(349)
Chapter 2 (The First Law of Thermodynamics)
Problem 1
From the first law of thermodynamics in canonical form (dU = dQ − dW ) and the energy
unit conversion between cal’s and ergs of
1cal = 4.185 107erg ,
we obtain
dW = 3.4 108 erg
dQ = 32 cal = 32(4.185 107 erg) = 1.339 109 erg .
(350)
(351)
From these we obtain for dU
dU = 1.339 109 − 3.4 108 = 9.9 108 erg = 23.8 cal
(352)
Problem 2
The work done by an isothermal expansion is given by Fermi Eq. 9
m
p1
V2
m
=
.
RT log
RT log
L=
M
p2
M
V1
(353)
m
Since M
is the number of moles of the gas under consideration for the specifics for this
problem we have that
5
L = (3mol)(8.314 J/molK)(0 + 273.15 K) log
= 3.48 103J = 831.54 cal .
3
Since we are considering an ideal gas whos internal energy is a function of temperature only
the first law of thermodynamics dU = dQ − dW simplifies since the expansion considered is
isothermal (dU = 0) to give
dQ = dW .
Because of this relationship, we have that the number of calories absorbed, or Q, is given by
Q = W = L which is computed above.
Problem 3 (a straight line transformation in the (V, p) plane)
For this problem we desire to compute the amount of work performed by the gas when it
undergoes a straight line transformation between (V0 , p0 ) and (V1 , p1 ) in the (V, p) plane. In
the specific case given in this problem the initial and final volumes (and temperatures) are
specified while the initial and final pressures must be computed from the equation of state
for an ideal gas. Using the ideal gas law to derive the corresponding pressure p0 we have
p0 =
8.314 J/molK 291 K 1 mol
= 1.15 105Pa .
−2
3
2.1 10 m
In the same way we find
p1 = 1.99 105Pa .
Since we have not derived an expression for the work done under a straight line transformation in the (V, p) plane we do so now and then use its result to evaluate this problem. The
work a gas performs under any transformation in the (V, p) plane is given by
Z
Z
W = L = p dV = p(V ) dV .
Where we have expressed the path in the (V, p) plane as a function of V explicitly with the
notation p(V ). When the path in the (V, p) plane is a line connecting the state (V0 , p0 ) to
(V1 , p1 ) can derive and explicit formula for the path by equating the slope at any point on
the line to the slope between the two end points as
p1 − p0
p(V ) − p0
=
.
V − V0
V1 − V0
On solving for p = p(V ) we obtain
p = p0 +
p1 − p0
(V − V0 ) .
V1 − V0
Thus the work the gas must do as it traverses this path is given by the integral
Z V1 p1 − p0
L=
p0 +
(V − V0 ) dV .
V1 − V0
V0
The algebra needed to perform this integration is
V
p1 − p0 (V − V0 )2 1
p1 − p0 (V1 − V0 )2
L = p0 (V1 − V0 ) +
= p0 (V1 − V0 ) + V1 − V0
V1 − V0
2
2
V0
i
h
p1 p0
1
−
= p0 (V1 − V0 ) + (p1 − p0 )(V1 − V0 ) = (V1 − V0 ) p0 +
2
2
2
1
=
(V1 − V0 )(p1 + p0 ) .
2
Now V1 − V0 < 0 and thus the work is negative and the environment performs work on
the system. Note also that the absolute value of the above expression is the area of the
triangle connecting the three states (V0 , p0 ), (V1 , p1 ), and (V1 , p0 ) and the rectangle beneath
this triangle. Evaluating the above with the given values of p and V gives
1
L = (1.27 − 2.1) 10−2m3 (1.99 + 1.15) 105Pa = −1.3 103 m3 Pa .
2
To calculate the heat absorbed by the system we remember that for an ideal gas U(T ) = CV T
and a diatomic gas specifically has CV = 52 R so we have the change in internal energy given
by ∆U = CV ∆T = CV (T1 − T0 ) or
5
∆U = CV (305 − 291) = CV (14 K) = (8.314J/molK)(14K)(1mol) = 2.9 102J .
2
From the first law of thermodynamics for small changes we have ∆Q = ∆U + ∆W and we
obtain for the heat absorbed by the system
∆Q = 2.9 102J + (−1.3 103 J) = −10.1 102 J = −1.01 1010 erg .
Where we used the unit conversion between ergs and Joules of 1J = 107 erg.
Problem 4
In this problem we have one mole of a diatomic gas undergoing an adiabatic volume expansion. If we define the initial volume to be V0 the final volume V1 is then 1.35V0. For an
adiabatic transformation we have
T V K−1 = constant .
(354)
Restricting this expression to connect the two states (T0 , V0 ) and (T1 , V1 ) we obtain
T0 V0K−1 = T1 V1K−1
(355)
Solving for the final temperature T1 gives
T1 = T0
V0
V1
K−1
.
(356)
We are given the initial temperature of T0 = 18C = 18 + 273.15K = 291.15K, the knowledge
1
.
that the gas is diatomic (so K = 75 ), and the fact that ratio of volumes is given by VV01 = 1.35
From this information we can compute that
V0
V1
K−1
= 0.507 .
(357)
With this we finally compute the final temperature as T1 = 291.15K (0.507) = 147.6 K.
Chapter 3 (The Second Law of Thermodynamics)
Problem 1
Referring to figure, XXX, we first characterize what we know along each path in the p-V
plane and then develop the necessary mathematics. First, paths BD and CA are adiabatic
therefore no head flows through them. Second, along AB and DC the temperature does not
change and since for an ideal gas the internal energy U is a function of temperature only
U = U(T ). Therefore along AB and DC dU = 0 so from the first law we have that dQ = dW
which implies Q = W .
We will solve this problem for the total work performed by the gas by computing the work
along each path in a clockwise fashion beginning with the path AB. To perform this calculation we will require the work performed by an ideal gas along an adiabatic expansion. This
will be calculated first and used in the calculations that follow.
We begin by deriving the work performed during an adiabatic transformation between two
general points in p − V space (p1 , V1 ) and (p2 , V2 ). Along an adiabatic transformation of an
ideal gas we have
pV K = constant ,
(358)
or anchoring the “constant” in the above expression to the point (p1 , V1 ) we have
pV k = p1 V1k
(359)
or solving for p = p(V ) we obtain
p = p1
V1
V
K
= p1
V
V1
−K
(360)
Thus the work between to points can be calculated in a straight forward manner as
Z V2
W12 =
pdV
V1
Z V2
p1
=
V −k dV
V1−k
V1
−k+1 V2
p1
V
=
V −k −k + 1 V1
1
=
p1 V1k
1−k
V2−k+1 − V1−k+1 .
(361)
We now begin with the calculations required for this particular problem. First the work
performed by our ideal gas along the isothermal path AB is given by
WAB = RT2 log(
VB
),
VA
(362)
which when evaluated, using the numbers given in the problem, gives the following
5
WAB = (8.314J/molK)(400K) log( ) = 5352.3J .
1
(363)
Note that this is also equal to the heat absorbed by our engine Q2 , at this state of the cycle.
Now along the path BD, since it is adiabatic we have in terms of variables of this problem
T2 VBK−1 = T1 VDK−1 .
(364)
Solving for VD we have
VD =
T2 K−1
V
T1 B
1/K−1
= VB
T2
T1
1/K−1
.
(365)
Using the numbers from the text, we get that
1/K−1
400
VD = 5liters
.
300
Now for a monotonic gas K = 5/3 giving the volume in the D state the value
VD = 7.69 liter .
Using equation 361 we have that for a monotonic gas
(6.65 105Pa)(5 10−3m3 )5/3 −3 3 1− 53
−3 3 1− 53
(7.69
10
m
)
−
(5
10
m
)
= 1.24 103J .
WBD =
1 − 53
For practice, the units in this expression work out as follows
2
Pa m5 (m3 )− 3 = Pa m5 m−2
= Pa m3
N 3
m
=
m2
= Nm = J
Along the isothermal path DC we have that pV = constant giving in terms of the variables
of this problem the expression
pD V D = pC V C ,
(366)
which when we solve for pC the expression
VD
7.69
5
pC = pD
= 4.32 10 Pa
= 2.158 106 Pa .
VC
1.539
(367)
Since CA is another adiabatic curve we have again that
T2 VAK−1 = T1 VCK−1 ,
or solving for VC the expression
1/K−1
1/K−1
T2
4
VC =
VA = 1 liter
= 1.539 liter .
T1
3
(368)
Then the work along this segment of the cycle is given by
WCA = RT1 log(
7.69
VD
) = (8.314J/molK)(300K) log(
) = 4.01 103J .
VC
1.539
(369)
Note that this equals the heat released by the process through this segment of the cycle or
Q1 . Finally, along the segment CA we have (using the expression for the work performed by
an adiabatic transformation derived above) the following
2.158 106Pa (1.539 10−3m3 )5/3 −3 3 −2/3
−3 3 − 23
(10
m
)
−
(1.539
10
m
)
LCA =
= −1.65 103 J
5
3
So the total work performed during the entire cycle is given by
Wtotal = WAB + WBD + WDC + WCA = 9.27 103J
(370)
Problem 2
First, convert the given temperatures into Kelvin as follows
T1 = 18 C = 291.15 K
T2 = 400 C = 673.15 K ,
(371)
(372)
then using Fermi Eq. 59 we have a maximum possible efficiency between these two temperatures of
291.15
T1
=1−
= 0.567 .
(373)
η =1−
T2
673.15
Problem 3
Converting the given temperatures into Kelvin we have
0 F = 255.92 K = T1
100 F = 310.92 K = T2 .
(374)
(375)
Then the minimum amount of work will be achieved when the engines operating cycle is
reversible. As such, from Fermi Eq. 60. we have
T2 − T1
L=
Q1 .
(376)
T1
With the heat extracted Q1 = 1 cal we obtain, in various units
L =
=
=
=
=
0.2149 Q1
0.2149 cal
0.2149 4.185 107erg
8.99 106 erg
0.899 J .
Where in the above conversions we have used the conversion that 1 erg = 10−7 J.
Chapter 4 (The Entropy)
Problem 1
We begin by recognizing that 1kg H2 O = 1000 gm H2 O and converting the given temperatures
to Kelvin as follows
T1 = 0 + 273.15 K = 273.15 K
T2 = 100 + 273.15 K = 373.15 K
For liquid’s I’ll make the assumption that Cp ≈ CV and is constant. As given in the book
we will take its value to be 1cal/gm. A change in entropy along a reversible path is given by
Z B
dQ
∆S =
.
(377)
T
A
From the first law we have dU = dQ − pdV and a constant volume transformation results
in dU = dQ = CV dT . Since along a constant volume transformation we have the heat ratio
defined by
dQ
= CV
(378)
dT V
Putting this expression into equation 377 results in
Z T2
CV dT
T2
∆S =
= CV ln( ) .
T
T1
T1
(379)
Which is an expression to be understood as per unit mass. Using the numbers provided for
this problem gives for the entropy change
373.15
)
273.15
= 311.9 cal/K = 311.9 (4.185 107erg/K)
= 1.305 1010erg/K
∆S = (1000 gm)(1cal/gmK) ln(
Problem 2
We are told that a body obeys the following equation of state
pV 1.2 = 109 T 1.1 ,
and asked to find the energy and entropy of such a system as a function of T and V . To
solve this problem in a more general setting consider a body that has an equation of state
given by
pV a = dT b
(380)
With a = 1.2, d = 109 , b = 1.1, and CV = 0.1 cal/deg for this problem. Now the units of V
are liters. Now consider U = U(V, T ) then one can show for a system performing p-V work
that
∂U
∂p
=T
− p,
∂V T
∂T V
which is a nice expression because once an equation of state is specified empirically one can
compute/evaluate
∂U
,
∂V T
by explicitly inserting the given equation of state. In our case
p=
dTb
,
Va
(381)
so the required derivative is given by
therefore
∂U
∂V
=
T
d b T b−1
,
Va
(382)
dbT b−1 dT b
dT b
−
=
(b
−
1)
= (b − 1)p .
Va
Va
Va
(383)
∂p
∂T
=
V
From this expression integrating U holding T fixed gives the following expression for U(V, T ),
in terms of an arbitrary function of temperature C1
U(V, T ) = (b − 1)dT b
V −a+1
+ C1 (T ) .
−a + 1
To evaluate C1 (T ) consider the definition of CV (evaluated at V0 = 100 liters)
1−b ∂Q
∂U
b−1 V
+ C ′ (T ) .
CV ≡
=
= (b − 1)dbT
1
∂T V
∂T V
1 − a V0
which gives for C1′ (T ) the following
CV′
(T ) = CV − (b − 1)dbT
b−1
V01−b
.
1−a
Integrating with respect to T we get
C1 (T ) = CV T − (b − 1)dT b
V01−b
+ C2
1−a
with C2 a constant. Then U(V, T ) is given by
U(V, T ) = CV T + (b − 1) d T b
(V 1−a − V01−a )
+ C2
1−a
(384)
To compute the entropy remember that for an reversible transformation we have
dS =
dU + pdV
1
1
dQ
=
= dU + p dV .
T
T
T
T
(385)
Using the expression for dU from differential calculus
∂U
∂U
dU =
dT +
dV ,
∂T V
∂V T
in the above expression for dU we obtain for dS
dbT b−1 1−a
1 dT b
1
1−a
b −a
CV + (b − 1)
(V
− V0 ) dT + (b − 1)dT V dV +
dV
dS =
T
1−a
T Va
or manipulating this expression we have
CV
dT
T
dT
= CV
T
dT
= CV
T
dT
= CV
T
dS =
(b1 )dbT b−2 1−a
1
1 dT b
(V
− V01−a )dT + (b − 1)dT bV −a dV +
dV
1−a
T
T Va
bdT b V −a
b(b − 1)T b−2 dV 1−a
(b − 1)dbV01−a T b−2
+
dV +
dT −
dT
T
1−a
1−a
(b − 1)dbV01−a T b−2
d2 T b dV 1−a
−
dT + bdT b−1 V −a dV +
dT
1−a
dT 2 1 − a
V 1−a
d2 (T b ) V 1−a
d(T b ) −a
− b(b − 1)d 0 T b−2 dT +
d
dT
+
d
V dV
1−a
dT 2 1 − a
dT
+
We can verify our algebra by checking if this is an exact differential as it must be. As such
we must have
∂
d(T b) −a
d2 (T b ) V 1−a
∂ CV
=
d
+ ...+
d
V
∂V T
dT 2 1 − a
∂T
dT
which becomes
d2(T b ) −a
d2(T b ) −a
V
=
d
V
(386)
dT 2
dT 2
which is a true statement. Therefore one can integrate dS if one can find an integrating
factor. Since dS is an exact differential there exists a function S(T, V ) such that
d2 T b V 1−a
CV
dV 1−a
∂S
=d 2
+
− b(b − 1) 0 T b−2
(387)
∂T V
dT 1 − a
T
1−a
d
and
The second expression gives that
∂S
∂V
S=d
=d
T
d(T b) −a
V .
dT
d(T b) V 1−a
+ F (T ) ,
dT 1 − a
(388)
(389)
for an arbitrary function F (T ). Taking the temperature derivative of the above gives
d2 (T b ) V −a
∂S
=d
+ F ′ (T )
2
∂T V
dT 1 − a
and setting this equal to equation 387 gives for F ′ (T )
1−a
CV
b−2 V0
− b(b − 1)T
d
F (T ) =
T
1−a
′
which can be integrated to give for F (T )
F (T ) = CV ln(T ) − bT b−1
V01−a
d + S0 .
1−a
In total we have for S(T, V ) the following
S(T, V ) = CV ln(T ) +
dbT b−1 1−a
(V
− V01−a ) + S0
1−a
(390)
Problem 3
From Clapeyron’s equation we have
λ
dp
=
dT
T (V2 − V1 )
(391)
Here V2 is the specific volume of the gas phase and V1 is the specific volume of liquid phase.
In Kelvin, the boiling temperature for ethyl alcohol is given by T = 351.45K. Assuming
V1 ≪ V2 and that the value of V2 for ethyl alcohol is approximately the same as for H2 O of
3
Clapeyron’s equation becomes
1677 cm
gm
dp
dT
=
J
855 gm
3
)
(351.45K)(1677 10−6 m
gm
J
= 1.45 103 3
m K
Pa
= 1.45 103
.
K
Chapter 5 (Thermodynamic Potentials)
Problem 1
WWX: I have not finished proofreading this section ... start this is page 11 from
the second set of scanned notes
The phase rule v = 2 + n − f , saturated solution and a solid dissolved in the in the substitute
have two phase liquid solution and solid (know this is correct page 86 gives examples of salt
in H2 0 and two components n = 2 ( a solid and a liquid component ).
v = 2+2−2 =2
(392)
therefore we can specify 2 variables T and p arbitrary. I would think that the correct answer
wold be T only. I know that increasing the temperature increases the solvability but am not
sure about the pressure)
Problem 2
I am told the amounts of H2 O and air. I’ll assume air contains N2 , O2 , and H2 only and
H2 O vapor, then f equals the number of phases (which is 2), and n equals the number of
components which is 4. We have that
v = 2+4−2 =4
But we are told the amount of H2 0. and air so we are told
(393)
Problem 3
Now
v(t) = v0 + v1 t + v2 t2
(394)
v0 = 924
v1 = 0.0015
v0 = 0.0000061
(395)
(396)
(397)
The units of T are centigrade, the units of V are volts, and the units of e are Columnbs,
from Page 96 we have
dQ = dU + dL = −eU(T ) + ev(T )
(398)
dv
but from the equation of Helmholtz u(T ) = v − T dT
, thus the equation of Helmholtz gives
the a functional form for the energy lost per unit charge.
u(T ) = v0 + v1 t + v2 t2 − T (v1 + 2v2 t)
= (v0 + T v1 ) + (v1 − 2v2 T )t + v2 t2
(399)
(400)
dv
dv
) + ev = eT dT
therefore we need u(T ) actually
Then dQ = −e(v − T dT
∆Q = eT
dv
= eT (v1 + 2v2 t)
dT
Chapter 6 (Gaseous Reactions)
Chapter 7 (The Thermodynamics of Dilute Solutions)
Chapter 8 (The Entropy Constant)
WWX: I have not finished this section ... end
(401)